Prove that $n\cdot(n-1)\cdot(n-2)\cdot\dots\cdot(n-r+1)=\dfrac{n!}{(n-r)!}$

Question

Prove that $n\cdot(n-1)\cdot(n-2)\cdot\dots\cdot(n-r+1)=\dfrac{n!}{(n-r)!}$

I know that $n\cdot(n-1)\cdot(n-2)\dots$ is $n!$, but how can I prove the denominator of $(n-r)!$?

Answer 1

$$(n-r)!=(n-r)(n-r-1)\cdots2\cdot1$$

$$n!=n(n-1)(n-2)\cdots(n-r+1)(n-r)(n-r-1)\cdots2\cdot1$$

Thus,

$$\frac{n!}{(n-r)!}=\frac{n(n-1)(n-2)\cdots(n-r+1)\overbrace{(n-r)(n-r-1)\cdots2\cdot1}}{\underbrace{(n-r)(n-r-1)\cdots2\cdot1}}\\\vphantom{\cfrac11}=n(n-1)(n-2)\cdots(n-r+1)$$

Answer 2

Hint:

$n! = n \cdot (n-1) \cdot (n-2) \cdot \cdots \cdot (n-r+1) \cdot (n-r) \cdot \cdots \cdot 2 \cdot 1$

Answer 3

multiply both the denominator and the numerator to $1\cdot 2\cdot 3\cdot ...\cdot \left( n-r \right) $ $$n\cdot (n-1)\cdot (n-2)\cdot \dots \cdot (n-r+1)=\frac { 1\cdot 2\cdot 3\cdot ...\cdot \left( n-r \right) \cdot (n-r+1)\cdot ...\cdot (n-2)\cdot (n-1)\cdot n }{ 1\cdot 2\cdot 3\cdot ...\cdot \left( n-r \right) } =\frac { n! }{ \left( n-r \right) ! } $$

Answer 4

$$n\cdot(n-1)\cdot(n-2)\cdot\dots\cdot(n-r+1)$$ Multiplying denominator and numerator by $$(n-r)(n-r-1)....(2)(1)$$

gives $$n\cdot(n-1)\cdot(n-2)\cdot\dots\cdot(n-r+1)\cdot\frac{(n-r)(n-r-1)....(2)(1)}{(n-r)(n-r-1)....(2)(1)}=\frac{n!}{(n-r)!}$$

Answer 5

Let's compute the number of $r$-permutations of $n$ in two different ways.

On one hand we can place one of $n$ objects to the first playce, one of $n - 1$ objects on the second place, etc. to one of $n - r + 1$ objects on the $r$th place. So there are $n \cdot (n - 1) \cdots (n - r + 1)$ of $r$-permutations of $n$.

On the other hand we can take each of $n!$ permutations of $n$ objects and selects first $r$ objects. Then each of $r$-permutations of $n$ will be achieved $(n - r)!$ times, once for each permutation of $n - r$ objects not being on the first $r$ positions. Thus there are $\frac{n!}{(n - r)!}$ of $r$-permutations of $n$.